Question

In each of the following problems, z represents the displacement of a particle from the origin. Find (as functions of t) its speed and the magnitude of its acceleration, and describe the motion.

$\mathit{z}\mathbf{=}\mathbf{(}\mathbf{1}\mathbf{+}\mathit{i}\mathbf{)}{\mathit{e}}^{\mathit{i}\mathit{t}}$

Short answer

The part it follows is circular$x^2+y^2=2$.

Its speed is $\left|v\right|=\sqrt{2}\mathrm{m}/\mathrm{s}$.

Its acceleration is $\left|A\right|=\sqrt{2}\mathrm{m}·{\mathrm{s}}^{-2}$.

Step-by-step solution

Given Information.

The given expression is, $z=\left(1+i\right)e^{it}$.

Meaning of rectangular form.

Represent the complex number in rectangular form means writing the given complex number in the form of $\mathit{x}\mathbf{+}\mathit{i}\mathit{y}\mathbf{}$in which x is the real part and y is the imaginary part.

Find the magnitude.

Write the general form of the complex number.

$\begin{array}{l}z=x+yI\\ z=a{e}^{\left(i\omega t\right)}\\ z=\left(1+i\right)e^{it}\end{array}$

Find the magnitude.

$\begin{array}{l}\left|z\right|=\sqrt{x^2+y^2}\\ \left|z\right|=\left|\left(1+i\right)e^{\left(it\right)}\right|\\ \left|z\right|=\left|\left(1+\mathrm{i}\right)\right|\\ \left|z\right|=\sqrt{2}\end{array}$

Therefore, the equation becomes$x^2+y^2=2$

Find the velocity and acceleration.

Differentiate with respect to time.

$\begin{array}{l}v=\frac{d}{dt}\left[(1+i)exp\left(it\right)\right]\\ v=i(1+i)exp\left(it\right)\end{array}$

Find the magnitude of the velocity.

$$\begin{gathered} \left|v\right|=\left|i\left(i+1\right)exp\left(it\right)\right| \\ \left|v\right|=\left|-1+i\right| \\ \left|v\right|=\sqrt{2}m/s \end{gathered}$$

Find the acceleration.

Differentiate the velocity with respect to time.

$\begin{array}{l}A=\frac{d}{dt}\left[i\left(1+i\right)e^{\left(it\right)}\right]\\ A=-\left(1+i\right)e^{\left(it\right)}\end{array}$

Find the magnitude of the acceleration.

$$\begin{gathered} \left|A\right|=\left|-(\mathrm{i}+1)\exp \left(\mathrm{it}\right)\right| \\ \left|A\right|=\left|1+i\right|\sqrt{2}{\mathrm{ms}}^{-2} \end{gathered}$$

Therefore, the path it follows it circular$x^2+y^2=2$.

Its speed is $\left|v\right|=\sqrt{2}\mathrm{m}/\mathrm{S}$.

Its acceleration is $\left|A\right|=\sqrt{2}{\mathrm{ms}}^{-2}$.